Let $g \geq 1$ be an integer, and $\mathrm{Sp}_{2g}(\mathbb{Z})$ be the symplectic group of $2g \times 2g$ integral matrices. In this case, we are defining the symplectic group a the automorphism group of a $2g$-dimensional vector space which preserves an alternating form $\langle \cdot, \cdot \rangle$, with respect to a symplectic basis $\{A_{1}, ... , A_{g}, B_{1}, ... , B_{g}\}$ such that $\langle A_{i}, A_{j} \rangle = \langle B_{i}, B_{j} \rangle = 0$, and $\langle A_{i}, B_{i} \rangle = -1$.

I am looking at subgroups of $\Gamma(2)$ generated by powers of these transvections. In particular, I want to know, given such a subgroup, what is the largest principal congruence subgroup it contains. There are several issues to worry about here:

1) It's not even clear to me how to determine when a subgroup generated by powers of these transvections is of finite index in $\Gamma(2)$.

2) Even if it is of finite index, when $g = 1$ not all subgroups of finite index are congruence subgroups, so in that case it may not contain a principal congruence subgroup at all. Although, what I really care about is what subgroup of $\mathrm{Sp}_{2g}(\mathbb{Z}_{2})$ they topologically generate, so this may not be an issue.

3) It may be possible to use software to compute individual examples, but what I'm mainly interested in is results on how to determine this for a general choice of powers of the $2g^{2} + g$ transvections listed above.

It would be greatly appreciated if anyone could point me to results that would be relevant in this situation (or a closely related one). Thanks!

The question of what subgroup of Sp(Z_2) they topologically generate is really different in flavor from (and much easier than) the question of describing the subgroup of Sp(Z) they generate.
–
JSEJul 4 '13 at 20:08

1 Answer
1

There is a Theorem of Tits in Comptes Rendus of Academy of Paris "Systemes generateurs grupes de congruence" ...(early 80's) where he proves this (the finiteness of index of the group generated by powers of the root group generators) for any Chevalley group of $Q$ rank at least two. For the symplectic group, the root group generators coincide with the generators that you have listed.